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Verify mean value theorem for each of the function $f(x)=\large\frac{1}{4x-1}$ in $[1,4].$

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  • Let $f(x)$ be a function defined on $[a,b]$ such that
  • (i) it is continuous on $[a,b]$
  • (ii) it is differentiable on $(a,b)$
  • Then,there exists $c\in (a,b)$ such that $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
Given : $f(x)=\large\frac{1}{4x-1}$ in $[1,4]$
Since for each $x\in [1,4]$,the function attains a unique definite value.
So $f(x)$ is continuous on $[1,4]$
Also $f'(x)=\large\frac{-4}{(4x-1)^2}$ exists for all $x\in (1,4)$.So it is differentiable on $(1,4)$.Hence both the conditions of Lagrange's Mean value theorem are satisfied.
Hence there must be one value of $c\in (1,4)$ such that $f'(c)=\large\frac{f(4)-f(1)}{4-1}$
Step 2:
Therefore $f'(c)=\large\frac{-4}{(4c-1)^2}$
$f(1)=\large\frac{1}{4\times 1-1}=\frac{1}{3}$
$f(4)=\large\frac{1}{4\times 4-1}=\frac{1}{15}$
Therefore $f'(x)=\large\frac{1/15-1/3}{4-1}=\frac{-4}{45}$
$\Rightarrow \large\frac{-4}{4(-1)^2}=\frac{-4}{45}$
$\Rightarrow 45=4(-1)^2$
$\Rightarrow 16c^2-8c+1=45$
On simplifying we get,
Step 3:
Solving for $c$ by applying the quadratic formula,
But $\sqrt{89}=9.43$
Neglecting the negative value
Clearly the value of $c$ lies between $[1,4]$
Hence Lagrange's Mean value theorem is verified.
answered Jul 3, 2013 by sreemathi.v
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